Use triple integrals to find the volume of the ellipsoid given by
$$4x^2+y^2+4z^2=16$$
ok never did these so clueless?
Hey bigwave.
You will need to consider a double integral [a suitable one] and find the appropriate limits or what is called a surface integral to relate an integral involving the boundary of the surface to the volume [which is the study of vector calculus].
I'd think about fixing one variable and then getting a function in terms of the others and once you do that find a way of getting the limits of integration that correspond to the volume you need to get and evaluate that integral.
You can do z as a function of (x,y), x as a function of (y,z) or y as a function of (x,z) but you need to make sure the limits only cover the region that is the ellipsoid.
I would first divide both sides by 16 to get the "standard form" $\frac{x^2}{4}+ \frac{y^2}{16}+ \frac{z^2}{4}= 1$. so this is an ellipsoid with center at (0, 0, 0), x and z going from -2 to 2 and y from -4 to 4. If we were to take the "outer integral" to be with respect to z then z goes from -2 to 2. For each z, then, $\frac{x^2}{4}+ \frac{y^2}{16}= 1- \frac{z^2}{4}= \frac{4- z^2}{4}$, an ellipse. Divide both sides by $\frac{4- z^2}{4}$ to get the standard form $\frac{x^2}{4- z^2}+ \frac{y^2}{4(4- z^2)}= 1$. That is an ellipse with center at (0, 0), x going from $-\sqrt{4- z^2}$ to $\sqrt{4- z^2}$ and y from $-2\sqrt{4- z^2}$ to $2\sqrt{4- z^2}$. finally, for each y and z, we have $\frac{x^2}{4}= 1- \frac{z^2}{4}- \frac{y^2}{16}$. x goes from $-2\sqrt{1- \frac{z^2}{4}- \frac{y^2}{16}}$ to $2\sqrt{1- \frac{z^2}{4}- \frac{y^2}{16}}$.
The volume is given by
$\int_{-2}^2\int_{-2\sqrt{4- z^2}}^{2\sqrt{4- z^2}}\int_{-2\sqrt{1- \frac{z^2}{4}- \frac{y^2}{16}}}^{2\sqrt{1- \frac{z^2}{4}- \frac{y^2}{16}}} dxdydz$.
You can simplify that by finding the volume of the part of the ellipsoid in the first quadrant then multiplying by 4:
$4\int_0^2\int_0^{2\sqrt{4- z^2}}\int_0^{2\sqrt{1- \frac{z^2}{4}- \frac{y^2}{16}}} dxdydz$.